# Thread: Arclength of linear curve

1. ## Arclength of linear curve

I'm to calculate the line segment y=mx+b between (x0, y0) and (x1, y1), using the formula for arclength when y is a function of x, that is:

L = (integral evaluated at a and b) (root)(1 + (f'(x))^2

Which in this case should give:

L = (integral evaluated at x0 and x1) (root) (1+m^2), as m is the derivative of mx+b.

This I get to be [(root)(1+m^2)*t]evaluated at x0 and x1, but the answer in my textbook is (1+m^2)|x0-x1|... any help would be appreciated! Thanks!

$\displaystyle \int_{x_0}^{x_1}\sqrt{1+m^2}dx$

$\displaystyle \sqrt{1+m^2}$ this is just some arbitrary number so move outside the integral.

$\displaystyle =\sqrt{1+m^2}*\int_{x_0}^{x_1}dx$

$\displaystyle =\sqrt{1+m^2}*\big[x\big]_{x_0}^{x_1}$

$\displaystyle =\sqrt{1+m^2}*(x_1-x_0)$

This what I obtained as a solution to the definite integral.

3. Originally Posted by gralla55
I'm to calculate the line segment y=mx+b between (x0, y0) and (x1, y1), using the formula for arclength when y is a function of x, that is:

L = (integral evaluated at a and b) (root)(1 + (f'(x))^2

Which in this case should give:

L = (integral evaluated at x0 and x1) (root) (1+m^2), as m is the derivative of mx+b.

This I get to be [(root)(1+m^2)*t]evaluated at x0 and x1, but the answer in my textbook is (1+m^2)|x0-x1|... any help would be appreciated! Thanks!
remember that you're setting up a general case arc length integral.

what if $\displaystyle x_1 < x_0$ ?

4. Skeeter the book also got rid of the radical unless he forgot to type it.

5. Yes, the missing root sign is what confused me... but I guess the book could be wrong. Thanks anyway!